250+ TOP MCQs on Efficiency of Synchronous Machines and Answers

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Electrical Machines Questions on “Efficiency of Synchronous Machines”.

1. The short circuit load losses is/are ______________
A. direct load loss and stray load losses
B. direct load loss
C. stray load losses
D. field current loss

Answer: A
Clarification: Short circuit losses comprise of direct load loss and stray losses too.

2. The open circuit voltage and open circuit core loss variation is?

Answer: A
Clarification: Core losses vary as a square of the voltage.

3. Rotational losses vary as following with respect to field current.

Answer: A
Clarification: The y-intercept depicts the friction and windage losses.

4. A 100 KVA, 400V, 3-phase, star connected alternator due to the following data:

Friction and windage losses = 340W
Open circuit core loss = 480W
Rf = 180ohms, ra = 0.02ohms.

The voltage applied to the field winding is 220V. The short circuit load loss at half full load is?
A. 258W
B. 268W
C. 480w
D. 340W

Answer: A
Clarification: SC losses at half load = 3*((Ia/2)2)*ra
= 3*((131.22/2)2)*0.02
= 258.3 A.

5. A 100 KVA, 400V, 3-phase, star connected alternator due to following data:

Friction and windage losses = 340W
Open circuit core loss = 480W
Rf = 180ohms, ra = 0.02ohms.

The voltage applied to the field winding is 220V. The field current loss will be?
A. 270W
B. 258W
C. 480W
D. 250W

Answer: A
Clarification: Field circuit loss = 220^2/180 = 169W

6. A 100 KVA, 400V, 3-phase, star connected alternator due to following data:

Friction and windage losses = 340W
Open circuit core loss = 480W
Rf = 180ohms, ra = 0.02ohms.

The voltage applied to the field winding is 220V. The efficiency of the machine at half load is?
A. 96.7%
B. 95%
C. 94.2%
D. 97

Answer: A
Clarification: Total losses at half load = 340+480+258+268.9 = 1347W
Efficiency = 1-(1347/40000+1347)
= 96.47%.

7. A 100 KVA, 400V, 3-phase, star connected alternator due to following data:

Friction and windage losses = 340W
Open circuit core loss = 480W
Rf = 180ohms, ra = 0.02ohms.

The voltage applied to the field winding is 220V. The load losses at full load is?
A. 1033W
B. 500W
C. 540W
D. 940W

Answer: A
Clarification: Short circuit losses at full load = 3*(131.22)2*0.02 = 1033W.

8. A 100 KVA, 400V, 3-phase, star connected alternator due to following data

Friction and windage losses = 340W
Open circuit core loss = 480W
Rf = 180ohms, ra = 0.02ohms.

The voltage applied to the field winding is 220V. Efficiency at full load is?
A. 97%
B. 96.5%
C. 92%
D. 95%

Answer: A
Clarification: Total losses at full load = 340+480+1033+268.9 = 2122W
Efficiency = 1 – (2122/2122+80000) = 97.4%.

9. The given variable losses are 5kW fr a 500 KVA, 11 kV, 3-phase star connected alternator having armature resistance of 4 ohms. Calculate the current at which maximum efficiency occurs?
A. 20.4A
B. 10.2A
C. 40A
D. 26A

Answer: A
Clarification: I2=5000/3*4
I = 20.4 A.

10. The given variable losses are 5kW fr a 500 KVA, 11 kV, 3-phase star connected alternator having armature resistance of 4 ohms. Calculate the full load armature current per phase.
A. 26.24 A
B. 20.41 A
C. 79 A
D. 40 A

Answer: A
Clarification: I = P/1.73*V = 5000/1.73*11000 = 26.24 A.

11. A synchronous machine with its field winding on stator and polyphase armature winding on rotor. At steady state, its air gap field is?

I. stationary w.r.t. stator 
II. rotating at double the speed Ns w.r.t. rotor
III. rotating in direction opposite to rotor

A. I, II, III
B. I
C. II, III
D. II, I

Answer: A
Clarification: All the statements are correct.

12. A synchronous machine with its field winding on rotor and polyphase armature winding on stator. At steady state running condition, its air gap field is?
A. rotating at synchronous speed w.r.t. stator
B. stationary w.r.t. rotor
C. rotating in the direction of the rotor rotor rotation
D. all of the mentioned

Answer: A
Clarification: As the field is on the rotor, the field will be set up w.r.t. rotor.

13. Consider a 3-phase cylindrical-rotor alternator.

A. E.m.f. generated by armature reaction lags armature current by 90°.
B. Air gap voltage leads the field flux by 90°
C. Air gap voltage lags the field flux by 90°    
D. Armature reaction mmf lags the field flux by (90°+ internal p.f. angle)

A. A, B, D
B. A, C, D
C. B, D
D. C, D

Answer: A
Clarification: Air gap voltage will lag the field flux by 90° to generate the emf.

14. In a 3-phase cylindrical-rotor alternator, synchronous reactance is sum of _____________
A. mutual and leakage reactance
B. magnetizing and leakage reactance
C. magnetizing and mutual reactance
D. mutual, magnetizing and leakage reactance

Answer: A
Clarification: Xs = magnetizing reactance + leakage reactance.

15. The reactive power output of a synchronous generator is limited by _________
A. armature current and field current
B. field current and load angle
C. load angle and excitation
D. armature current only

Answer: A
Clarification: Q=f(Ia,If).